A Diophantine Equation | x^y=y^x

#Shorts #numbertheory #equations #multiplication #maths #math #mathematics via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

[ZantierTasa]

If anybody is wondering why these steps seemed to avoid the answer x=y, it's because you need to be careful with the steps you take in maths.

One of the steps was "raise both sides to the power of 1/(k-1)" But this is not a valid operation for k=1. So to be correct, you would break this into cases:

case 1: k != 1. We get the solutions in the video.
case 2: k = 1. In this case, x=y, and we can see from the initial equation that this is a valid solution for x, y != 0.

[FrumiousBandersnatch]

So I tried to plug this into Desmos and looked at the geometric location of all the points that can be expressed as (t^(1/t-1), t^(t/t-1)). Then I put in the original equation x^y=y^x.
Naturally, the equation overlaps with the geometric location except for all the points for which x=y ("the generic case"). However, the geometric location has two additional bits that aren't in the positive-positive quarter. What are those and where do they come from? What are all the points (t^(1/t-1), t^(t/t-1)) that aren't on the equation x^y=y^x?

[aidancoletta]

There is another reason that x=y isn't allowed. Plug the new x and y values ( in terms of k) back into the initial formula and use logs to reduce. You get that k+1/k-1=1, which if you notice, has no solution since no k will satisfy that equation. That was very disappointing.

[Petro]

How do you get the substitution x=ky? X and Y could also be related differently and not linearly

[spelunkerd]

Some how, the most obvious solution, x=y, is not allowed with a solution formula that produces zero in a denominator.